2022-05-22：给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
百度百科中最近公共祖先的定义为：“对于有根树 T 的两个节点 p、q，最近公共祖先表示为一个节点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。
如何时间复杂度O(N)，额外空间复杂度O(1)，解决最低公共祖先问题?
力扣236。二叉树的最近公共祖先。
答案2022-05-22：
莫里斯遍历。
答案用rust编写，答案有误。代码如下：
代码语言：rust复制use std::cell::RefCell;
use std::rc::Rc;
fn main() {
let mut head = Rc::new(RefCell::new(Some(TreeNode::new(3))));
let mut left = Rc::clone(&head.borrow().as_ref().unwrap().left);
*left.borrow_mut() = Some(TreeNode::new(5));
let mut right = Rc::clone(&head.borrow().as_ref().unwrap().right);
*right.borrow_mut() = Some(TreeNode::new(1));
let mut head2 = Rc::clone(&head.borrow().as_ref().unwrap().left);
let mut leftleft = Rc::clone(&head2.borrow().as_ref().unwrap().left);
*leftleft.borrow_mut() = Some(TreeNode::new(6));
let mut rightright = Rc::clone(&head2.borrow().as_ref().unwrap().right);
*rightright.borrow_mut() = Some(TreeNode::new(2));
let ans = lowest_common_ancestor(
Rc::clone(&head),
Rc::clone(&leftleft),
Rc::clone(&rightright),
);
if ans.borrow().is_none() {
println!("None");
} else {
println!("ans = {}", ans.borrow().as_ref().unwrap().val);
}
let mut left = Rc::clone(&head.borrow().as_ref().unwrap().left);
let mut right = Rc::clone(&head.borrow().as_ref().unwrap().right);
println!("head = {}", head.borrow().as_ref().unwrap().val);
println!("p = {}", left.borrow().as_ref().unwrap().val);
println!("q = {}", right.borrow().as_ref().unwrap().val);
println!("---------------");
println!("head = {}", head.borrow().as_ref().unwrap().val);
println!("left = {}", left.borrow().as_ref().unwrap().val);
println!("right = {}", right.borrow().as_ref().unwrap().val);
println!("left.left = {}", leftleft.borrow().as_ref().unwrap().val);
println!("left.right = {}", rightright.borrow().as_ref().unwrap().val);
}
pub struct TreeNode {
pub val: i32,
pub left: Rc<RefCell<Option<TreeNode>>>,
pub right: Rc<RefCell<Option<TreeNode>>>,
}
impl TreeNode {
pub fn new(val: i32) -> Self {
Self {
val,
left: Rc::new(RefCell::new(None)),
right: Rc::new(RefCell::new(None)),
}
}
}
fn lowest_common_ancestor(
mut head: Rc<RefCell<Option<TreeNode>>>,
mut o1: Rc<RefCell<Option<TreeNode>>>,
mut o2: Rc<RefCell<Option<TreeNode>>>,
) -> Rc<RefCell<Option<TreeNode>>> {
if !find_first(
Rc::clone(&o1.borrow().as_ref().unwrap().left),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
|| !find_first(
Rc::clone(&o1.borrow().as_ref().unwrap().right),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
{
return Rc::clone(&o1);
}
if !find_first(
Rc::clone(&o2.borrow().as_ref().unwrap().left),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
|| !find_first(
Rc::clone(&o2.borrow().as_ref().unwrap().right),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
{
return Rc::clone(&o1);
}
let mut left_aim: Rc<RefCell<Option<TreeNode>>> =
find_first(Rc::clone(&head), Rc::clone(&o1), Rc::clone(&o2));
let mut cur: Rc<RefCell<Option<TreeNode>>> = Rc::clone(&head);
let mut most_right: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
let mut ans: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
while cur.borrow().is_none() {
most_right = Rc::clone(&cur.borrow().as_ref().unwrap().left);
if !most_right.borrow().is_none() {
while !Rc::clone(&most_right.borrow().as_ref().unwrap().right)
.borrow()
.is_none()
&& is_eq(
Rc::clone(&most_right.borrow().as_ref().unwrap().right),
Rc::clone(&cur),
)
{
let mut mostrightright = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
most_right = Rc::clone(&mostrightright);
}
if Rc::clone(&most_right.borrow().as_ref().unwrap().right)
.borrow()
.is_none()
{
let mut mostrightright = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
mostrightright = Rc::clone(&cur);
let mut curleft = Rc::clone(&cur.borrow().as_ref().unwrap().left);
cur = Rc::clone(&curleft);
continue;
} else {
let mut mostrightright = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
mostrightright = Rc::new(RefCell::new(None));
if find_left_aim(
Rc::clone(&cur.borrow().as_ref().unwrap().left),
Rc::clone(&left_aim),
) {
if ans.borrow().is_none()
&& !find_first(
Rc::clone(&left_aim.borrow().as_ref().unwrap().right),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
{
ans = Rc::clone(&left_aim);
}
left_aim = Rc::clone(&cur);
}
}
}
let mut curright = Rc::clone(&cur.borrow().as_ref().unwrap().right);
cur = Rc::clone(&curright);
}
return if !ans.borrow().is_none() {
ans
} else {
if !find_first(
Rc::clone(&left_aim.borrow().as_ref().unwrap().right),
Rc::clone(&o1),
Rc::clone(&o2),
)
.borrow()
.is_none()
{
Rc::clone(&left_aim)
} else {
Rc::clone(&head)
}
};
}
fn is_eq(mut a: Rc<RefCell<Option<TreeNode>>>, mut b: Rc<RefCell<Option<TreeNode>>>) -> bool {
if a.borrow().is_none() && b.borrow().is_none() {
return true;
}
if a.borrow().is_none() || b.borrow().is_none() {
return false;
}
return a.borrow().as_ref().unwrap().val == b.borrow().as_ref().unwrap().val;
}
fn find_left_aim(
mut head: Rc<RefCell<Option<TreeNode>>>,
mut left_aim: Rc<RefCell<Option<TreeNode>>>,
) -> bool {
let mut tail = reverse_edge(head);
let mut cur = Rc::clone(&tail);
let mut ans = false;
while !cur.borrow().is_none() {
if is_eq(Rc::clone(&cur), Rc::clone(&left_aim)) {
ans = true;
}
let mut curright = Rc::clone(&cur.borrow().as_ref().unwrap().right);
cur = Rc::clone(&curright);
}
reverse_edge(tail);
return ans;
}
fn reverse_edge(mut from: Rc<RefCell<Option<TreeNode>>>) -> Rc<RefCell<Option<TreeNode>>> {
let mut pre: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
let mut next: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
while !from.borrow().is_none() {
next = Rc::clone(&from.borrow().as_ref().unwrap().right);
{
let mut fromright = Rc::clone(&from.borrow().as_ref().unwrap().right);
fromright = Rc::clone(&pre); //此处错误
}
pre = Rc::clone(&from);
from = Rc::clone(&next);
}
return pre;
}
fn find_first(
mut head: Rc<RefCell<Option<TreeNode>>>,
mut o1: Rc<RefCell<Option<TreeNode>>>,
mut o2: Rc<RefCell<Option<TreeNode>>>,
) -> Rc<RefCell<Option<TreeNode>>> {
if head.borrow().is_none() {
return Rc::new(RefCell::new(None));
}
let mut cur: Rc<RefCell<Option<TreeNode>>> = Rc::clone(&head);
let mut most_right: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
let mut first: Rc<RefCell<Option<TreeNode>>> = Rc::new(RefCell::new(None));
while !cur.borrow().is_none() {
most_right = Rc::clone(&cur.borrow().as_ref().unwrap().left);
if !most_right.borrow().is_none() {
while !Rc::clone(&most_right.borrow().as_ref().unwrap().right)
.borrow()
.is_none()
&& !is_eq(
Rc::clone(&most_right.borrow().as_ref().unwrap().right),
Rc::clone(&cur),
)
{
let mut most_right_right = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
most_right = Rc::clone(&most_right_right);
}
if Rc::clone(&most_right.borrow().as_ref().unwrap().right)
.borrow()
.is_none()
{
if !first.borrow().is_none()
&& (is_eq(Rc::clone(&cur), Rc::clone(&o1))
|| is_eq(Rc::clone(&cur), Rc::clone(&o2)))
{
first = Rc::clone(&cur);
}
let mut most_right_right = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
most_right_right = Rc::clone(&cur); //此处错误
let mut curleft = Rc::clone(&cur.borrow().as_ref().unwrap().left);
cur = Rc::clone(&curleft);
continue;
} else {
let mut most_right_right = Rc::clone(&most_right.borrow().as_ref().unwrap().right);
most_right_right = Rc::new(RefCell::new(None)); //此处错误
}
} else {
if first.borrow().is_none()
&& (is_eq(Rc::clone(&cur), Rc::clone(&o1))
|| is_eq(Rc::clone(&cur), Rc::clone(&o2)))
{
first = Rc::clone(&cur);
}
first = Rc::clone(&cur);
}
let mut curright = Rc::clone(&cur.borrow().as_ref().unwrap().right);
cur = Rc::clone(&curright);
}
return first;
}执行结果如下：
在这里插入图片描述答案用golang编写。代码如下：
代码语言：go复制package main
import "fmt"
func main() {
head := &TreeNode{val: 3}
head.left = &TreeNode{val: 5}
head.right = &TreeNode{val: 1}
head.left.left = &TreeNode{val: 6}
head.left.right = &TreeNode{val: 2}
ans := lowestCommonAncestor(head, head.left.left, head.left.right)
fmt.Println(ans.val)
}
// Definition for a binary tree node.
type TreeNode struct {
val int
left *TreeNode
right *TreeNode
}
// 提交以下的方法
// 该方法亮点在于：时间复杂度O(N)，额外空间复杂度O(1)
func lowestCommonAncestor(head, o1, o2 *TreeNode) *TreeNode {
if findFirst(o1.left, o1, o2) != nil || findFirst(o1.right, o1, o2) != nil {
return o1
}
if findFirst(o2.left, o1, o2) != nil || findFirst(o2.right, o1, o2) != nil {
return o2
}
leftAim := findFirst(head, o1, o2)
cur := head
var mostRight *TreeNode
var ans *TreeNode
for cur != nil {
mostRight = cur.left
if mostRight != nil {
for mostRight.right != nil && mostRight.right != cur {
mostRight = mostRight.right
}
if mostRight.right == nil {
mostRight.right = cur
cur = cur.left
continue
} else {
mostRight.right = nil
if findLeftAim(cur.left, leftAim) {
if ans == nil && findFirst(leftAim.right, o1, o2) != nil {
ans = leftAim
}
leftAim = cur
}
}
}
cur = cur.right
}
if ans != nil {
return ans
} else {
if findFirst(leftAim.right, o1, o2) != nil {
return leftAim
} else {
return head
}
}
}
func findLeftAim(head, leftAim *TreeNode) bool {
tail := reverseEdge(head)
cur := tail
ans := false
for cur != nil {
if cur == leftAim {
ans = true
}
cur = cur.right
}
reverseEdge(tail)
return ans
}
func reverseEdge(from *TreeNode) *TreeNode {
var pre *TreeNode
var next *TreeNode
for from != nil {
next = from.right
from.right = pre
pre = from
from = next
}
return pre
}
func findFirst(head, o1, o2 *TreeNode) *TreeNode {
if head == nil {
return nil
}
cur := head
var mostRight *TreeNode
var first *TreeNode
for cur != nil {
mostRight = cur.left
if mostRight != nil {
for mostRight.right != nil && mostRight.right != cur {
mostRight = mostRight.right
}
if mostRight.right == nil {
if first == nil && (cur == o1 || cur == o2) {
first = cur
}
mostRight.right = cur
cur = cur.left
continue
} else {
mostRight.right = nil
}
} else {
if first == nil && (cur == o1 || cur == o2) {
first = cur
}
}
cur = cur.right
}
return first
}执行结果如下：
在这里插入图片描述左神java代码
